Fourier Series of modulus[t] - example of this?

[ZedCar]

Homework Statement

Does anyone know of a website, or a book, where I can see a worked example of the Fourier Series

f(t) = [t]
-∏<t<∏
T=2∏

Finding a0 and an

Of course, it doesn't have to be t, it could be x or any other variable.

Thank you.

Homework Equations

The Attempt at a Solution

 

[ZedCar]

I've found an example here;
http://www.exampleproblems.com/wiki/index.php/FS1

Why is it that in my notes 'an' = -4/[∏(k^2)]

 

[OldEngr63]

You have defined the function as
f(t) = [t]
-∏<t<∏
T=2∏
but that is not the example that you show worked out. The example that you show worked out is this one
f(t) = |t|
-∏<t<∏
T=2∏
This difference being that, in the latter case, the function between - π and +π is the absolute value of t; in the original problem statement, as you gave it, [t] is not really a well defined function.

The absolute value function is an even function, so when we proceed to evaluate the coefficients for the Fourier series, the first one works out as

a_o = [itex]\frac{1}{2π}[/itex]∫_-π^π f(t) dt
= [itex]\frac{1}{π}[/itex]∫₀^π t dt
=[itex]\frac{π}{2}[/itex]

The others follow in the usual fashion with the cosine multiplication.

 

[vela]

I've found an example here;
http://www.exampleproblems.com/wiki/index.php/FS1

Why is it that in my notes 'an' = -4/[∏(k^2)]

That should be either n or k, not both, in your equation. Let's just use n.

What does the result of the integration give you when you assume n is odd and n is even?

 

[paranormal]

The Fourier Series is a mathematical tool used to represent a periodic function as a sum of sine and cosine functions. In this case, the function is f(t) = [t] over the interval -∏<t<∏ with a period of T=2∏. The modulus function, denoted by [t], returns the absolute value of t, meaning that it will always be positive.

To find the Fourier Series of this function, we first need to find the coefficients a0 and an. The coefficient a0 is given by the formula a0 = (1/T) ∫f(t)dt over one period. In this case, we have T=2∏, so a0 = (1/2∏) ∫[t]dt = 1/∏.

The coefficients an are given by the formula an = (2/T) ∫f(t)cos(nωt)dt over one period, where ω = 2∏/T. In this case, we have ω = 2∏/(2∏) = 1, so an = (2/2∏) ∫[t]cos(nt)dt = (1/∏) ∫[t]cos(nt)dt.

To find this integral, we can use integration by parts. Let u = [t] and dv = cos(nt)dt. Then du = sign(t)dt and v = (1/n)sin(nt). By the integration by parts formula, we have ∫uv = uv - ∫vu. Plugging in our values, we get ∫[t]cos(nt)dt = [t](1/n)sin(nt) - ∫(1/n)sin(nt)sign(t)dt. Since the function is periodic, we only need to evaluate this integral over one period, which is from -∏ to ∏.

Using the same substitution, we get ∫(1/n)sin(nt)sign(t)dt = -(1/n^2)cos(nt)sign(t) + ∫(1/n^2)cos(nt)dt. Evaluating this integral over one period, we get ∫(1/n^2)cos(nt)dt = 0, since cos(nt) is an odd function over one period.

Therefore, an = (1/∏) ∫[t]cos(nt